Question Number 196525 by Mastermind last updated on 26/Aug/23
Answered by mr W last updated on 27/Aug/23
$$\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}>\sqrt{{m}^{\mathrm{2}} }={m} \\ $$$$\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}<\sqrt{{m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{1}}=\sqrt{\left({m}+\mathrm{1}\right)^{\mathrm{2}} }={m}+\mathrm{1} \\ $$$${since}\:{m}<\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}<{m}+\mathrm{1}, \\ $$$$\Rightarrow\left[\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right]={m} \\ $$
Commented by Mastermind last updated on 27/Aug/23
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$